【leetcode】Palindrome Partitioning

Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [

    ["aa","b"],

    ["a","a","b"]

  ]

为了加速运算，可以利用动态规划，求出满足回文的子串的位置

 

palindrome[i][j]表示了第字符串中，s[i,i+1,……, j]是否是回文

 

可以有以下递推公式：

if(i==j) palindrome[i][j]=true;

if(i-j=1)palindrome[i][j]=s[i]==s[j];

if(i-j>1)palindrome[i][j]=palindrome[i+1][j-1]&&s[i]==s[j]

 

 

得到了该回文表后，我们利用回溯法得到所有的子串

 

 

 

 1 class Solution {

 2 public:

 3  

 4     vector<vector <string> > res;

 5    

 6     vector<vector<bool> > palindrome;

 7     string s;

 8     int n;

 9    

10     vector<vector<string>> partition(string s) {

11        

12          this->s=s;

13          this->n=s.length();

14          

15          vector<vector<bool> > palindrome(n,vector<bool>(n));

16          getPalindrome(palindrome);

17          this->palindrome=palindrome;

18          

19          vector <string> tmp;

20          getPartition(0,tmp);

21          

22          return res;

23     }

24    

25     //回溯得到子串

26     void getPartition(int start,vector<string> tmp)

27     {

28  

29         if(start==n)

30         {

31             res.push_back(tmp);

32             return;

33         }

34        

35         for(int i=start;i<n;i++)

36         {

37             if(palindrome[start][i])

38             {

39                 tmp.push_back(s.substr(start,i-start+1));

40                 getPartition(i+1,tmp);

41                 tmp.pop_back();

42             }

43         }

44     }

45    

46    

47    

48     void getPalindrome(vector<vector<bool> > &palindrome)

49     {

50         int startIndex=0;

51         int endIndex=n-1;

52        

53         for(int i=n-1;i>=0;i--)

54         {

55             for(int j=i;j<n;j++)

56             {

57                 if(i==j)

58                 {

59                     palindrome[i][j]=true;

60                 }

61                 else if(j-i==1)

62                 {

63                     palindrome[i][j]=(s[i]==s[j]);

64                 }

65                 else if(j-i>1)

66                 {

67                     palindrome[i][j]=(s[i]==s[j]&&palindrome[i+1][j-1]);

68                 }

69             }

70         }

71     }

72 };